Which one of the following is **TRUE**?

From the given table, we calculate the Need matrix ($ext{Need} = ext{Max} - ext{Allocation}$):

• $\text{Need}(P_0) = (8 - 0,\; 4 - 0,\; 3 - 1) = (8, 4, 2)$

• $\text{Need}(P_1) = (6 - 3,\; 2 - 2,\; 0 - 0) = (3, 0, 0)$

• $\text{Need}(P_2) = (3 - 2,\; 3 - 1,\; 3 - 1) = (1, 2, 2)$

Currently, $\text{Available} = (3, 2, 2)$.

---

### Analyzing REQ1: $P_0$ requests $(0, 0, 2)$.

• Request $\le \text{Need}(P_0)$ and Request $\le \text{Available}$.

• If granted tentatively:

  • $\text{Allocation}(P_0) = (0, 0, 3)$

  • $\text{Need}(P_0) = (8, 4, 0)$

  • $\text{Available} = (3, 2, 2) - (0, 0, 2) = (3, 2, 0)$

Now, test for a safe sequence:

• $$\text{Need}$(P_0) = (8, 4, 0)

leq (3, 2, 0)$

• $\text{Need}(P_1) = (3, 0, 0) \le (3, 2, 0) \implies P_1$ finishes, new $\text{Available} = (3, 2, 0) + (3, 2, 0) = (6, 4, 0)$

• Now:

  • $$\text{Need}$(P_0) = (8, 4, 0)

leq (6, 4, 0)$-$$\text{Need}$(P_2) = (1, 2, 2) leq (6, 4, 0)$(since$Z$has$0$available but requires$2$)

• Neither $P_0$ nor $P_2$ can proceed. Hence, the system enters an unsafe state.

• Therefore, REQ1 cannot be permitted.

---

### Analyzing REQ2: $P_1$ requests $(2, 0, 0)$.

• Request $\le \text{Need}(P_1)$ and Request $\le \text{Available}$.

• If granted tentatively:

  • $\text{Allocation}(P_1) = (5, 2, 0)$

  • $\text{Need}(P_1) = (1, 0, 0)$

  • $\text{Available} = (3, 2, 2) - (2, 0, 0) = (1, 2, 2)$

Now, test for a safe sequence:

• $\text{Need}(P_1) = (1, 0, 0) \le (1, 2, 2) \implies P_1$ finishes, new $\text{Available} = (1, 2, 2) + (5, 2, 0) = (6, 4, 2)$

• Next, $\text{Need}(P_2) = (1, 2, 2) \le (6, 4, 2) \implies P_2$ finishes, new $\text{Available} = (6, 4, 2) + (2, 1, 1) = (8, 5, 3)$

• Next, $\text{Need}(P_0) = (8, 4, 2) \le (8, 5, 3) \implies P_0$ finishes.

• A safe sequence $\langle P_1, P_2, P_0 \rangle$ exists.

• Therefore, REQ2 can be permitted.

Thus, Only REQ2 can be permitted (Option B).